# binomial confidence interval from multiple observations

As far as I understand, a confidence interval for a population proportion (e.g. p = probability of a coin coming up heads) is constructed based on one observation of X=k, X~ binomial (n,p) (let's say k heads observed in n flips).

I am now thinking that if you have more data, e.g. k1, k2, k3 from the same distribution X ~ binomial (n,p), you have more information so you should be able to construct a narrower interval. Is that true, and if so, what is the method for doing that?

• What do you mean by ", a confidence interval for the binomial p" – Kane Chua Nov 10 '19 at 4:20
• Can you clarify: Are you thinking of a case where e.g. you have a bag of n balls that are each either red, green, or blue, so that p1= blue/n, and so on? – Sal Mangiafico Nov 10 '19 at 15:00

If $$X_1, X_2, X_3$$ are independent and identically distributed as $$\mathsf{Binom}(n, p),$$ then $$S = X_1 + X_2 + X_3 \sim \mathsf{Binom}(m=3n, p).$$

So you can use $$\hat p = S/m$$ as the unbiased point estimate of $$p.$$ Then you can use your formula for a binomial confidence interval (CI) to get an interval estimate. Use $$S$$ and $$m$$ instead of $$X$$ and $$n.$$ A CI based on $$3n$$ observations will tend to be shorter than a CI based on $$n$$ observations (but not 1/3 the length).

I will not show the CI because several styles of binomial CIs are in common use and you do not indicate which one you are using.

If I understand your question correctly, you make $$n$$ independent trials, but split these up into $$n=n_1+n_2+n_3$$ and compute $$k_1$$, $$k_2$$, $$k_3$$. As BruceET already mentioned this is not a good idea, because you use your data in an inefficient way and obtain too large confidence intervals.

I can only guess what your reasoning might be behind splitting up the trials; maybe it is based on biology textbooks which, according to this ComputationlSciene question recommend this approach for error estimation. The idea behind this suggestion is to use the different results for cross-validation, but there are always better methods (see the discussion of the above question). In this particular case, this is pointless, because many good confidence intervals are known for a binomial $$p$$, see e.g.

L. D. Brown, T. T. Cai, and A. DasGupta, “Interval estimation for a binomial proportion,”Statistical science, vol. 16, no. 2, pp. 101–117, 2001

Brown et al. recommend the Wilson interval, but if $$p$$ is close to one or zero, the Bayesian HPD interval has even better coverage probability. It can only be computed numerically, though. So if you need a closed formualy, go for the Wilson interval.